Successfully reported this slideshow.
We use your LinkedIn profile and activity data to personalize ads and to show you more relevant ads. You can change your ad preferences anytime.
Solutions to Worksheet for Section 5.5
                             Integration by Substitution
                          ...
cos 3x dx
  4.                  , u = 5 + 2 sin 3x
         5 + 2 sin 3x

       Solution. We have du = 6 cos 3x dx, so
  ...
1
     Solution. Let u = 3x3 − 1, so du = 9x2 dx and dx =             du. So
                                             ...
Solution. In the first substitution, du = 2z dz and the integral becomes
                               du                 ...
12.   Find
                                              sec x dx

by multiplying the numerator and denominator by sec x +...
Upcoming SlideShare
Loading in …5
×

Lesson 28: Integration by Substitution (worksheet solutions)

3,006 views

Published on

  • Login to see the comments

  • Be the first to like this

Lesson 28: Integration by Substitution (worksheet solutions)

  1. 1. Solutions to Worksheet for Section 5.5 Integration by Substitution V63.0121, Calculus I Summer 2010 1. (3x − 5)17 dx, u = 3x − 5 1 Solution. As suggested, we let u = 3x − 5. Then du = 3 dx so dx = du. Thus 3 1 1 1 18 1 (3x − 5)17 dx = u17 du = · u +C = (3x − 5)18 + C 3 3 18 54 4 2. x x2 + 9 dx, u = x2 + 9 0 Solution. We have du = 2x dx, which takes care of the x and dx that appear in the integrand. The limits are changed to u(0) = 9 and u(4) = 25. Thus 4 25 1 25 √ 1 2 x x2 + 9 dx = u du = · u3/2 0 2 9 2 3 9 1 98 = · (125 − 27) = . 3 3 √ e x √ 3. √ dx, u = x. x 1 Solution. We have du = √ dx, which means that 2 x √ e x √ dx = 2 eu du = 2eu + C x 1
  2. 2. cos 3x dx 4. , u = 5 + 2 sin 3x 5 + 2 sin 3x Solution. We have du = 6 cos 3x dx, so cos 3x dx 1 du 1 1 = = ln |u| + C = ln |5 + 2 sin 3x| + C 5 + 2 sin 3x 6 u 6 6 In these problems, you need to determine the substitution yourself. 5. (4 − 3x)7 dx. 1 Solution. Let u = 4 − 3x, so du = −3x dx and dx = − du. Thus 3 1 u8 1 (4 − 3x)7 dx = − u7 du = − + C = − (4 − 3x)8 + C 3 24 24 π/3 6. csc2 (5x) dx π/4 1 Solution. Let u = 5x, so du = 5 dx and dx = du. So 5 π/3 5π/3 1 csc2 (5x) dx = csc2 u du π/4 5 5π/4 5π/3 1 =− cot u 5 5π/4 1 5π 5π = cot − cot 5 4 3 √ 1 3 = 1+ 5 3 3 −1 7. x2 e3x dx 2
  3. 3. 1 Solution. Let u = 3x3 − 1, so du = 9x2 dx and dx = du. So 9 3 −1 1 1 x2 e3x dx = eu du = eu + C 9 9 1 3x3 −1 = e . 9 Sometimes there is more than one way to skin a cat: x 8. Find dx, both by long division and by substituting u = 1 + x. 1+x Solution. Long division yields x 1 =1− 1+x 1+x So x dx dx = x − 1+x 1+x To find the leftover integral, let u = 1 + x. Then du = dx and so dx du = = ln |u| + C 1+x u Therefore x dx = x − ln |x + 1| + C 1+x Making the substitution immediately gives du = dx and x = u − 1. So x u−1 1 dx = du = 1− du 1+x u u = u − ln |u| + C = x + 1 − ln |x + 1| + C It may appear that the two solutions are “different.” But the difference is a constant, and we know that antiderivatives are only unique up to addition of a constant. 2z dz , both by substituting u = z 2 + 1 and u = 3 9. Find √ 3 z 2 + 1. z2 + 1 3
  4. 4. Solution. In the first substitution, du = 2z dz and the integral becomes du 3 2/3 3 √ = u−1/3 du = u + C = (z 2 + 1)2/3 3 u 2 2 In the second, u3 = z 2 + 1 and 3u2 du = 2z dz. The integral becomes 3u2 3 3 3 du = 3u2 du = u + C = (z 2 + 1)2/3 + C. u 2 2 The second one is a dirtier substitution, but the integration is cleaner. Use the trigonometric identity cos 2α = cos2 α − sin2 α = 2 cos2 α − 1 = 1 − 2 sin2 α to find 10. sin2 x dx Solution. Using cos 2α = 1 − 2 sin2 α, we get 1 − cos 2α sin2 α = 2 So 1 1 sin2 x dx = − cos 2x dx 2 2 x 1 = − sin 2x + C. 2 4 11. cos2 x dx Solution. Using cos 2α = 2 cos2 α − 1, we get 1 + cos 2α sin2 α = 2 So 1 1 sin2 x dx = + cos 2x dx 2 2 x 1 = + sin 2x + C. 2 4 4
  5. 5. 12. Find sec x dx by multiplying the numerator and denominator by sec x + tan x. Solution. We have sec x(sec x + tan x) sec x dx = dx sec x + tan x sec2 x + sec x tan x = dx tan x + sec x Now notice the numerator is the derivative of the denominator. So the substitution u = tan x+sec x gives sec x dx = ln |sec x + tan x| + C. 5

×